01 · Downward V
Find the vertex, range and zeros of y = 5 − |2x + 4|.
Hint
The bars vanish at x = −2.
Worked solution
Vertex (−2, 5), range y ≤ 5. Zeros satisfy 2x + 4 = ±5, giving x = 1/2 or x = −9/2.
Understand · explore · practise
Modulus graphs and problem solving
Solve modulus graph problems with parameters, restricted inverses, inequalities and areas. Distinguish zero, one, two and infinitely many intersections using an interactive model.
Before you startModulus equations, inequalities, graph transformations and inverse functions
01 / Read a transformed V
Find its vertex and range before solving intersections.
y = 2|x − 3| − 1Worked example
Vertex: (3, −1)
The expression inside the bars vanishes at x = 3.
Range: y ≥ −1
The multiplier 2 is positive, so the V opens upwards.
Left rule: y = 5 − 2x for x ≤ 3
Right rule: y = 2x − 7 for x ≥ 3
The branches agree at their shared vertex.
Zeros: x = 5/2 and x = 7/2
Solve |x − 3| = 1/2.
A negative outside multiplier turns the V downwards. For y = 4 − 3|x + 2|, the vertex is (−2, 4) and the range is y ≤ 4.
02 / Restrict before reversing
A full V is many-to-one; either complete branch can be inverted.
For f(x) = 2|x − 3| − 1, choose the domain x ≥ 3. On that branch f(x) = 2x − 7, giving f⁻¹(x) = (x + 7)/2 with domain x ≥ −1 and range y ≥ 3.
Choosing x ≤ 3 instead gives the branch f(x) = 5 − 2x. Its inverse is (5 − x)/2, with domain x ≥ −1 and range y ≤ 3. The original branch determines which inverse is valid.
03 / Intersect with a line
Each candidate must lie on the branch used to obtain it.
To solve |x| = mx + k, use the right and left rays separately. Count a valid vertex solution only once.
Right ray x ≥ 0: (1 − m)x = k
Left ray x ≤ 0: (−1 − m)x = k
Solve |x| = ½x + 3Worked example
Right: x = ½x + 3 ⇒ x = 6
This lies on the right ray.
Left: −x = ½x + 3 ⇒ x = −2
This lies on the left ray.
Intersections: (−2, 2) and (6, 6)
Both original equations are satisfied.
If a coefficient becomes zero, do not divide by it. A branch equation 0x = 0 holds on the whole branch, whereas 0x = k with k ≠ 0 has no solution there.
04 / Count roots with parameters
Parallel and coincident rays are special cases.
For |x| = mx + k, the following classification counts distinct real solutions on the full graph.
- |m| < 1: two if k > 0, one if k = 0, none if k < 0.
- |m| = 1: one if k > 0, infinitely many if k = 0, none if k < 0.
- |m| > 1: exactly one for every real k.
At m = 1 and k = 0, the line y = x coincides with the entire right ray. At m = −1 and k = 0, it coincides with the entire left ray. A meeting at the vertex therefore need not mean exactly one solution.
Count actual intersectionsExplore
Blue: y = |x|. Gold: y = mx + k. Green marks valid intersections; a green ray means infinitely many, continuing beyond the picture.
Equation: |x| = 0x + 1.
2 solutions: x = −1, 1.
Check the complete rays, not just whether the line meets the vertex.
Watch a single intersection become a coincident ray
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Use the intersection intervals
A sign chart or branch calculation decides which side works.
Solve |x| ≤ ½x + 3Worked example
For x ≥ 0: x ≤ ½x + 3 ⇒ x ≤ 6
Together with x ≥ 0, this gives [0, 6].
For x ≤ 0: −x ≤ ½x + 3 ⇒ x ≥ −2
Together with x ≤ 0, this gives [−2, 0].
Combine: −2 ≤ x ≤ 6
Equality includes both boundary inputs. A strict < would exclude −2 and 6.
For the inequality |x| ≥ ½x + 3, the answer is x ≤ −2 or x ≥ 6. The point x = 0 is an easy check: 0 ≥ 3 is false, so the middle interval cannot be included.
06 / A region between a V and a line
Find the vertices before choosing an area method.
The region |x| ≤ y ≤ ½x + 3 is a triangle with vertices O = (0, 0), A = (−2, 2) and B = (6, 6). The upper boundary is the line segment AB; the lower boundary follows the two rays OA and OB.
Split at the y-axisWorked example
The upper line meets the y-axis at C = (0, 3)
OC is a vertical segment of length 3 inside the triangle.
Left triangle OAC: ½ × 3 × 2 = 3
The perpendicular distance from A to the y-axis is 2.
Right triangle OCB: ½ × 3 × 6 = 9
The perpendicular distance from B to the y-axis is 6.
Total area = 12 square units
Do not use the horizontal distance AB as a base with an unrelated vertical height.
More generally, for y = mx + k above y = |x| with |m| < 1 and k > 0, the enclosed triangular area is k²/(1 − m²). This follows by adding two triangles with shared vertical base k and horizontal heights k/(1 + m) and k/(1 − m).
07 / A quadratic inside the bars
A different base graph can produce more than two roots.
For |x² − 4| = k, no roots exist when k < 0. For k ≥ 0, solve x² − 4 = k and x² − 4 = −k, retaining only real roots and counting duplicates once.
Count distinct rootsWorked example
k = 0: x = ±2
The two branch equations coincide; there are two distinct roots.
0 < k < 4: x = ±√(4 + k), ±√(4 − k)
All four roots are real and distinct.
k = 4: x = ±√8 and x = 0
The zero root from ±√0 counts once, giving three roots.
k > 4: x = ±√(4 + k)
The equation x² = 4 − k has no real roots; two remain.
Do not apply the zero/one/two-root classification for a V to every graph containing modulus bars.
08 / Your turn
Draw enough of the graph to identify every branch.
Use exact values and count distinct solutions.
02 · Restricted inverse
f(x) = 3|x + 1| + 2 for x ≥ −1. Find f⁻¹ and its domain.
Hint
Use the right-hand branch before rearranging.
Worked solution
f(x) = 3x + 5 on this domain. Its inverse is (x − 5)/3 with domain x ≥ 2 and range y ≥ −1.
03 · Two intersections
Solve |x| = −½x + 3.
Hint
Use each ray and check the sign of the candidate input.
Worked solution
The right ray gives x = 2; the left gives x = −6. Both are valid.
04 · Coincident ray
Solve |x| = −x. Explain why the answer is not just x = 0.
Hint
On which half of the line is |x| equal to −x?
Worked solution
Every x ≤ 0 is a solution. The line and the left ray coincide, so there are infinitely many solutions.
05 · A parallel line
How many solutions does |x| = x − 2 have?
Hint
On the right ray the equation becomes 0 = −2.
Worked solution
None. The left calculation gives x = 1, which is not on the left ray. Both branches therefore fail.
06 · Steeper line
Solve |x| = 2x − 4.
Hint
The line can cross one ray even with a negative intercept.
Worked solution
The right ray gives x = 4, valid. The left gives x = 4/3, rejected. The only solution is 4.
07 · Inequality
Solve |x| < −½x + 3.
Hint
The intersection inputs are −6 and 2; test the middle.
Worked solution
The solution is −6 < x < 2. The endpoints are excluded by the strict sign.
08 · Area
Find the area of the region |x| ≤ y ≤ 4.
Hint
The vertices are (−4, 4), (0, 0) and (4, 4).
Worked solution
The horizontal base has length 8 and perpendicular height 4. Area = ½ × 8 × 4 = 16 square units.
09 · A parameter count
For |x| = mx + 2, which real m give two distinct solutions?
Hint
The intercept is positive. Compare the line slope with both ray slopes.
Worked solution
Exactly −1 < m < 1. At m = ±1 there is one solution; for |m| > 1 there is also one.
10 · Four, three or two?
How many distinct real roots does |x² − 9| = k have when (a) k = 5, (b) k = 9 and (c) k = 12?
Hint
Solve x² = 9 + k and x² = 9 − k.
Worked solution
(a) Four: ±√14 and ±2. (b) Three: ±√18 and 0. (c) Two: ±√21; the other branch asks for x² = −3.
09 / Recap
Count valid inputs on the full graph, including exceptional cases.
- Start from the vertex, branches and range.
- Restrict a V to one branch before finding an inverse.
- Keep branch conditions beside parameter equations.
- A zero coefficient may mean an entire coincident ray.
- Use intersections to solve inequalities and find geometric regions.
- A quadratic inside modulus bars can have up to four roots.
Section 1 of 9 · Read a transformed V